Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Probability Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- Two coins are tossed. What is the probability of getting atleast one head?
- \(\frac{4}{3}\)
- \(\frac{3}{4}\)
- \(\frac{1}{2}\)
- \(\frac{2}{3}\)

Answer: (b) \(\frac{3}{4}\)

Solution: total number of events = \(4\)

Probability that no head occurs = \(\frac{1}{4}\)

then probability that atleast one head occurs $$ = 1 - \frac{1}{4} = \frac{3}{4} $$

Solution: total number of events = \(4\)

Probability that no head occurs = \(\frac{1}{4}\)

then probability that atleast one head occurs $$ = 1 - \frac{1}{4} = \frac{3}{4} $$

- From the pack of \(52\) playing cards, two cards are picked at random. What is the probability that both the cards are kings?
- \(\frac{1}{22}\)
- \(\frac{1}{220}\)
- \(\frac{1}{221}\)
- \(\frac{1}{222}\)

Answer: (c) \(\frac{1}{221}\)

Solution: total number of cards = \(52\)

number of kings = \(4\), then $$ P = \frac{^4C_2}{^{52}C_2}....(1) $$ where as $$ ^4C_2 = \frac{4 \times 3}{2} = 6 $$ $$ ^{52}C_2 = \frac{52 \times 51}{2} = 1326 $$ by putting these values in equation \((1)\), $$ P = \frac{6}{1326} = \frac{1}{221} $$

Solution: total number of cards = \(52\)

number of kings = \(4\), then $$ P = \frac{^4C_2}{^{52}C_2}....(1) $$ where as $$ ^4C_2 = \frac{4 \times 3}{2} = 6 $$ $$ ^{52}C_2 = \frac{52 \times 51}{2} = 1326 $$ by putting these values in equation \((1)\), $$ P = \frac{6}{1326} = \frac{1}{221} $$

- Three coins are tossed. What is the probability of getting two heads?
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{5}\)

Answer: (a) \(\frac{1}{2}\)

Solution: sample space = \(\{HHH, \ HHT, \ HTT, \ HTH\}\) \(\{TTT, \ THH, \ TTH, THT\}\)

total number of events = \(2^3\) = \(8\)

number of favourable outcomes = \(4\)

then probability of getting two heads, $$ P = \frac{4}{8} = \frac{1}{2} $$

Solution: sample space = \(\{HHH, \ HHT, \ HTT, \ HTH\}\) \(\{TTT, \ THH, \ TTH, THT\}\)

total number of events = \(2^3\) = \(8\)

number of favourable outcomes = \(4\)

then probability of getting two heads, $$ P = \frac{4}{8} = \frac{1}{2} $$

- If five coins are thrown, then find the probability of getting atleast one tail?
- \(\frac{32}{31}\)
- \(\frac{31}{32}\)
- \(\frac{1}{32}\)
- \(\frac{1}{31}\)

Answer: (b) \(\frac{31}{32}\)

Solution: total number of events = \(2^5\) = \(32\)

probability that no tail occurs = \(\frac{1}{32}\)

then probability that atleast one tail occurs, $$ = 1 - \frac{1}{32} = \frac{31}{32} $$

Solution: total number of events = \(2^5\) = \(32\)

probability that no tail occurs = \(\frac{1}{32}\)

then probability that atleast one tail occurs, $$ = 1 - \frac{1}{32} = \frac{31}{32} $$

- A bag contains \(6\) white, \(7\) black and \(8\) Red balls. If a ball is drawn at random, then what is the probability that it is not black?
- \(\frac{2}{3}\)
- \(\frac{1}{3}\)
- \(\frac{1}{2}\)
- \(\frac{3}{4}\)

Answer: (a) \(\frac{2}{3}\)

Solution: the probability that the ball is of black colour = \(\frac{7}{21}\) = \(\frac{1}{3}\)

then the probability that the ball is not of black colour = \(1 - \frac{1}{3}\) = \(\frac{2}{3}\)

Solution: the probability that the ball is of black colour = \(\frac{7}{21}\) = \(\frac{1}{3}\)

then the probability that the ball is not of black colour = \(1 - \frac{1}{3}\) = \(\frac{2}{3}\)

- If three coins are thrown, then find the probability of not getting head?
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{6}\)
- \(\frac{1}{8}\)

Answer: (d) \(\frac{1}{8}\)

Solution: total number of possible outcomes = \(8\)

there is only one way of not getting head, so

favourable outcome = \(1\), then $$ P = \frac{1}{8} $$

Solution: total number of possible outcomes = \(8\)

there is only one way of not getting head, so

favourable outcome = \(1\), then $$ P = \frac{1}{8} $$

- A bag contains \(15\) Red and \(18\) Green balls. If one ball is drawn at random, then what is the probability that it is not Red?
- \(\frac{6}{12}\)
- \(\frac{6}{13}\)
- \(\frac{6}{11}\)
- \(\frac{6}{10}\)

Answer: (c) \(\frac{6}{11}\)

Solution: the probability that the ball is of Red colour = \(\frac{15}{33}\) = \(\frac{5}{11}\)

the probability that the ball is not of Red colour, $$ = 1 - \frac{5}{11} = \frac{6}{11} $$

Solution: the probability that the ball is of Red colour = \(\frac{15}{33}\) = \(\frac{5}{11}\)

the probability that the ball is not of Red colour, $$ = 1 - \frac{5}{11} = \frac{6}{11} $$

- A fair dice is thrown, What is the probability of getting a odd number?
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{5}\)

Answer: (a) \(\frac{1}{2}\)

Solution: number of dice = \(\{1, \ 2, \ 3, \ 4, \ 5, \ 6\}\)

odd numbers on dice = \(\{1, \ 3, \ 5\}\)

number of possible outcomes = \(6\)

number of favourable outcomes = \(3\), then $$ P = \frac{3}{6} = \frac{1}{2} $$

Solution: number of dice = \(\{1, \ 2, \ 3, \ 4, \ 5, \ 6\}\)

odd numbers on dice = \(\{1, \ 3, \ 5\}\)

number of possible outcomes = \(6\)

number of favourable outcomes = \(3\), then $$ P = \frac{3}{6} = \frac{1}{2} $$

- A bag contains \(3\) Red, \(4\) Black balls and an other bag contains \(5\) Red, \(6\) Black balls. If one ball is drawn from each bag, then find the probability that both are Red?
- \(\frac{12}{43}\)
- \(\frac{17}{67}\)
- \(\frac{15}{77}\)
- \(\frac{13}{57}\)

Answer: (c) \(\frac{15}{77}\)

Solution: Probability of getting both of Red colours, $$ P = \frac{3}{7} \times \frac{5}{11} $$ $$ P = \frac{15}{77} $$

Solution: Probability of getting both of Red colours, $$ P = \frac{3}{7} \times \frac{5}{11} $$ $$ P = \frac{15}{77} $$

- A bag contains \(5\) white and \(8\) Black balls. If three balls from the bag are drawn at random, then what is the probability that they are all white?
- \(\frac{5}{141}\)
- \(\frac{141}{5}\)
- \(\frac{5}{143}\)
- \(\frac{143}{5}\)

Answer: (c) \(\frac{5}{143}\)

Solution: total number of White balls = \(5\)

number of Black balls = \(8\), then $$ P = \frac{^5C_3}{^{13}C_3}....(1) $$ where as, $$ ^5C_3 = \frac{5 \times 4 \times 3}{3} = 20 $$ $$ ^{13}C_3 = \frac{13 \times 12 \times 11}{3} = 572 $$ by putting these values in equation \((1)\), we get $$ P = \frac{20}{572} = \frac{5}{143} $$

Solution: total number of White balls = \(5\)

number of Black balls = \(8\), then $$ P = \frac{^5C_3}{^{13}C_3}....(1) $$ where as, $$ ^5C_3 = \frac{5 \times 4 \times 3}{3} = 20 $$ $$ ^{13}C_3 = \frac{13 \times 12 \times 11}{3} = 572 $$ by putting these values in equation \((1)\), we get $$ P = \frac{20}{572} = \frac{5}{143} $$

Lec 1: Introduction to Probability
Exercise-1
Lec 2: Addition Rule
Exercise-2
Lec 3: Multiplication Rule
Exercise-3
Exercise-4
Exercise-5